Simplify and expand the following expression: $ \dfrac{q + 9}{q + 2}-\dfrac{q}{3q + 2} $
In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(q + 2)(3q + 2)$ Multiply the first term by $\dfrac{3q + 2}{3q + 2}$ $ \begin{align*} \dfrac{q + 9}{q + 2} \times \dfrac{3q + 2}{3q + 2} & = \dfrac{(q + 9)(3q + 2)}{(q + 2)(3q + 2)} \\ & = \dfrac{3q^2 + 29q + 18}{(q + 2)(3q + 2)}\end{align*} $ Multiply the second term by $\dfrac{q + 2}{q + 2}$ $ \begin{align*} \dfrac{q}{3q + 2} \times \dfrac{q + 2}{q + 2} & = \dfrac{(q)(q + 2)}{(3q + 2)(q + 2)} \\ & = \dfrac{q^2 + 2q}{(3q + 2)(q + 2)}\end{align*} $ Now we have: $ = \dfrac{3q^2 + 29q + 18}{(q + 2)(3q + 2)} - \dfrac{q^2 + 2q}{(3q + 2)(q + 2)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{3q^2 + 29q + 18 - (q^2 + 2q)}{(q + 2)(3q + 2)} $ $ = \dfrac{3q^2 + 29q + 18 - q^2 - 2q}{(q + 2)(3q + 2)} $ $ = \dfrac{2q^2 + 27q + 18}{(q + 2)(3q + 2)}$ Expand the denominator: $ = \dfrac{2q^2 + 27q + 18}{3q^2 + 8q + 4}$